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In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. $$ H_{0}: p_{1}=p_{2}=p_{3}=p_{4}=0.25 ; \quad n=500 $$

Short Answer

Expert verified
The expected counts for each category are 125.

Step by step solution

01

Understanding the null hypothesis

From the null hypothesis \(H_{0}: p_{1}=p_{2}=p_{3}=p_{4}=0.25\), it is given that the probability for each category is equal, i.e., 0.25.
02

Applying the formula for expected counts

The formula for expected counts is \(Expected Count = Total Sample Size * Probability\). In this case, the total sample size (n) is 500 and the probability (for each category due to the null hypothesis) is 0.25. Therefore, applying the formula for each category we get: \(Expected Count = 500 * 0.25 = 125\). Hence, the expected counts for each category are 125.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test
The Chi-Square Test is a popular statistical method used to determine if there’s a significant difference between expected and observed data. In the context of the exercise, we used the expected counts for each category, determined by the null hypothesis, to see if the actual observed data varies significantly from what's expected. The test involves comparing expected counts, calculated under a hypothesis, with observed counts from your data.
  • First, you calculate expected counts using your understanding of probabilities and sample size.
  • Then, you compare them with the actual counts you observe in your data.
This comparison helps determine if any observed deviation is due to randomness or if it suggests that the initial hypothesis may not be accurate. The Chi-Square Test is most effective when data is in categorical form, and it's essential for assumptions such as independence and uniformity to hold.
Probability
Probability is the mathematical measurement of the likelihood of an event occurring. In the given exercise, the null hypothesis states that the probability for each category is 0.25. This is a key aspect as it sets the groundwork for calculating expected counts. By assigning equal probabilities, the hypothesis assumes that no category is more likely than another to occur. Using probability in this exercise involves:
  • Understanding the probability assigned to each category based on the null hypothesis: \( p_1 = p_2 = p_3 = p_4 = 0.25 \).
  • Applying the uniform probability to calculate expected counts.
  • Comparing expected probabilities with observed probabilities to assess the accuracy of the hypothesis.
Probability provides a foundation for predicting outcomes and is crucial in determining whether observed data aligns with theoretical expectations.
Sample Size
Sample size is a critical factor that greatly influences the results and reliability of statistical analysis. In the exercise provided, the sample size of 500 is used to calculate the expected counts under the assumption that probabilities of all categories are equal. Understanding sample size involves:
  • Recognizing that a larger sample size can provide more accurate and reliable results.
  • Using the sample size to calculate expected counts, as shown in the formula: \[ \text{Expected Count} = \text{Sample Size} \times \text{Probability} \]
  • Knowing that changes in sample size can impact the outcome of a Chi-Square Test, as smaller samples might not capture the necessary variation.
Calculating expected counts accurately hinges on using the sample size and probability together. Thus, the sample size plays a pivotal role in chi-square calculations, influencing the confidence in your hypothesis testing.

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Most popular questions from this chapter

Give a two-way table and specify a particular cell for that table. In each case find the expected count for that cell and the contribution to the chi- square statistic for that cell. \((\) Group \(2,\) No \()\) $$ \begin{array}{l|rr} \hline & \text { Yes } & \text { No } \\ \hline \text { Group 1 } & 720 & 280 \\ \text { Group 2 } & 1180 & 320 \\ \hline \end{array} $$

In Exercises 7.5 to 7.8 , the categories of a categorical variable are given along with the observed counts from a sample. The expected counts from a null hypothesis are given in parentheses. Compute the \(\chi^{2}\) -test statistic, and use the \(\chi^{2}\) -distribution to find the p-value of the test. $$ \begin{array}{l} \hline \begin{array}{l} \text { Category } \\ \text { Observed } \\ \text { (Expected) } \end{array} & \begin{array}{c} \mathrm{A} \\ 38(30) \end{array} & \begin{array}{c} \mathrm{B} \\ 55(60) \end{array} & \begin{array}{c} \mathrm{C} \\ 79(90) \end{array} & \begin{array}{c} \mathrm{D} \\ 128(120) \end{array} \\ \hline \end{array} $$

In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. $$ \begin{aligned} &\mathbf{7 . 3} \quad H_{0}: p_{A}=0.50, p_{B}=0.25, p_{C}=0.25 ;\\\ &n=200 \end{aligned} $$

In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. $$ \begin{aligned} &\text { 7.4 } H_{0}: p_{1}=0.7, p_{2}=0.1, p_{3}=0.1, p_{4}=0.1 ;\\\ &n=400 \end{aligned} $$

Gender and Frequency of Status Updates on Facebook Exercise 7.48 introduces a study about users of social networking sites such as Facebook. Table 7.36 shows the self-reported frequency of status updates on Facebook by gender. Are frequency of status updates and gender related? Show all details of the test. $$ \begin{array}{l|rr|r} \hline \text { IStatus/Gender } \rightarrow & \text { Male } & \text { Female } & \text { Total } \\ \hline \text { Every day } & 42 & 88 & 130 \\ \text { 3-5 days/week } & 46 & 59 & 105 \\ \text { 1-2 days/week } & 70 & 79 & 149 \\ \text { Every few weeks } & 77 & 79 & 156 \\ \text { Less often } & 151 & 186 & 337 \\ \hline \text { Total } & 386 & 491 & 877 \\ \hline \end{array} $$

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